Java检索括号中的字符串字符

Question

I have a String in Java that holds data in this format:

String x = "xxxxxx xxxxxx xxxxxx (xxx)";

How can I set the value of that string to just be the characters in the brackets, without including the brackets? (Note that the character sizes will vary in each instance).

Answer 1

The one-line solution is to use String.replaceAll() and the appropriate regex that captures the whole input (effectively replacing the whole input), but also captures (non-greedily) the part you want as group 1, then puts back just that group:

String part = x.replaceAll(".*\\((.*?)\\).*", "$1"); 

FYI, the double back-slashes in the regex String is a single slash in regex, which then escapes the literal brackets in the regex

Here's some test code:

public static void main(String[] args) {
    String x = "xxxxxx xxxxxx xxxxxx (xxx)";
    String part = x.replaceAll(".*\\((.*)\\).*", "$1");
    System.out.println(part);
}

Output:

xxx    

Answer 2

A regex will do that, but my regex-fu is weak. The non-regex way is as follows:

int firstIndex = x.indexOf("(");
x = x.substring(firstIndex+1, x.length()-1);

EDIT: As pointed out in comments, if there are any other parentheses in the data, other than at the end, this will NOT work. You'd need to use the following instead:

int firstIndex = x.lastIndexOf("(", x.length()-6);
x = x.substring(firstIndex+1, x.length()-1);

EDIT2: Just reread and realised that the close paren is the last character. So there's no need to get the second index.

Answer 3

You can use String.split method for this kind of extraction..

String x = "xxxxxx xxxxxx xxxxxx (xxx)";    
String[] arr = x.split("\\(");

x = arr[1].substring(0, arr[1].indexOf(")")); // To remove the trailing bracket.

System.out.println(x);